B = μ 0 I 2 π R B = \frac{\mu_0I}{2\pi R} B = 2 π R μ 0 I
μ 0 ≈ 1.2567 ∗ 1 0 − 6 H m \mu_0\approx1.2567*10^{-6}\frac{H}{m} μ 0 ≈ 1.2567 ∗ 1 0 − 6 m H
B x = μ 0 I x 2 π R x B_x = \frac{\mu_0I_x}{2\pi R_x} B x = 2 π R x μ 0 I x
I x = 5 A I_x= 5A I x = 5 A
R x = 6 c m = 0.06 m distance to the X-axis R_x= 6cm =0.06m \text{ distance to the X-axis} R x = 6 c m = 0.06 m distance to the X-axis
B x = μ 0 ∗ 5 2 π ∗ 0.06 ≈ 1.67 ∗ 1 0 − 5 T B_x = \frac{\mu_0*5}{2\pi *0.06}\approx1.67*10^{-5}T B x = 2 π ∗ 0.06 μ 0 ∗ 5 ≈ 1.67 ∗ 1 0 − 5 T
B y = μ 0 I y 2 π R y B_y = \frac{\mu_0I_y}{2\pi R_y} B y = 2 π R y μ 0 I y
I y = 7 A I_y= 7A I y = 7 A
R y = 4 c m = 0.04 m distance to the Y-axis R_y= 4cm =0.04m \text{ distance to the Y-axis} R y = 4 c m = 0.04 m distance to the Y-axis
B y = μ 0 ∗ 7 2 π ∗ 0.04 ≈ 3.5 ∗ 1 0 − 5 T B_y = \frac{\mu_0*7}{2\pi *0.04}\approx3.5*10^{-5}T B y = 2 π ∗ 0.04 μ 0 ∗ 7 ≈ 3.5 ∗ 1 0 − 5 T
since the axes X ,Y are perpendicular \text{since the axes X ,Y are perpendicular} since the axes X ,Y are perpendicular
B = B x 2 + B y 2 = 1 0 − 5 1.6 7 2 + 3. 5 2 ≈ 3.88 ∗ 1 0 − 5 T B =\sqrt{B_x^2+B_y^2}= 10^{-5}\sqrt{1.67^2+3.5^2}\approx3.88*10^{-5}T B = B x 2 + B y 2 = 1 0 − 5 1.6 7 2 + 3. 5 2 ≈ 3.88 ∗ 1 0 − 5 T
Answer: B = 3.88 ∗ 1 0 − 5 T \text{Answer:}B =3.88*10^{-5}T Answer: B = 3.88 ∗ 1 0 − 5 T
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